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8x^{2}+15x+6=0
All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.
x=\frac{-15±\sqrt{15^{2}-4\times 8\times 6}}{2\times 8}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, 15 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-15±\sqrt{225-4\times 8\times 6}}{2\times 8}
Square 15.
x=\frac{-15±\sqrt{225-32\times 6}}{2\times 8}
Multiply -4 times 8.
x=\frac{-15±\sqrt{225-192}}{2\times 8}
Multiply -32 times 6.
x=\frac{-15±\sqrt{33}}{2\times 8}
Add 225 to -192.
x=\frac{-15±\sqrt{33}}{16}
Multiply 2 times 8.
x=\frac{\sqrt{33}-15}{16}
Now solve the equation x=\frac{-15±\sqrt{33}}{16} when ± is plus. Add -15 to \sqrt{33}.
x=\frac{-\sqrt{33}-15}{16}
Now solve the equation x=\frac{-15±\sqrt{33}}{16} when ± is minus. Subtract \sqrt{33} from -15.
x=\frac{\sqrt{33}-15}{16} x=\frac{-\sqrt{33}-15}{16}
The equation is now solved.
8x^{2}+15x+6=0
Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.
8x^{2}+15x+6-6=-6
Subtract 6 from both sides of the equation.
8x^{2}+15x=-6
Subtracting 6 from itself leaves 0.
\frac{8x^{2}+15x}{8}=-\frac{6}{8}
Divide both sides by 8.
x^{2}+\frac{15}{8}x=-\frac{6}{8}
Dividing by 8 undoes the multiplication by 8.
x^{2}+\frac{15}{8}x=-\frac{3}{4}
Reduce the fraction \frac{-6}{8} to lowest terms by extracting and canceling out 2.
x^{2}+\frac{15}{8}x+\left(\frac{15}{16}\right)^{2}=-\frac{3}{4}+\left(\frac{15}{16}\right)^{2}
Divide \frac{15}{8}, the coefficient of the x term, by 2 to get \frac{15}{16}. Then add the square of \frac{15}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{15}{8}x+\frac{225}{256}=-\frac{3}{4}+\frac{225}{256}
Square \frac{15}{16} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{15}{8}x+\frac{225}{256}=\frac{33}{256}
Add -\frac{3}{4} to \frac{225}{256} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{15}{16}\right)^{2}=\frac{33}{256}
Factor x^{2}+\frac{15}{8}x+\frac{225}{256}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.
\sqrt{\left(x+\frac{15}{16}\right)^{2}}=\sqrt{\frac{33}{256}}
Take the square root of both sides of the equation.
x+\frac{15}{16}=\frac{\sqrt{33}}{16} x+\frac{15}{16}=-\frac{\sqrt{33}}{16}
Simplify.
x=\frac{\sqrt{33}-15}{16} x=\frac{-\sqrt{33}-15}{16}
Subtract \frac{15}{16} from both sides of the equation.
x ^ 2 +\frac{15}{8}x +\frac{3}{4} = 0
Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 8
r + s = -\frac{15}{8} rs = \frac{3}{4}
Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C
r = -\frac{15}{16} - u s = -\frac{15}{16} + u
Two numbers r and s sum up to -\frac{15}{8} exactly when the average of the two numbers is \frac{1}{2}*-\frac{15}{8} = -\frac{15}{16}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>
(-\frac{15}{16} - u) (-\frac{15}{16} + u) = \frac{3}{4}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{3}{4}
\frac{225}{256} - u^2 = \frac{3}{4}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{3}{4}-\frac{225}{256} = -\frac{33}{256}
Simplify the expression by subtracting \frac{225}{256} on both sides
u^2 = \frac{33}{256} u = \pm\sqrt{\frac{33}{256}} = \pm \frac{\sqrt{33}}{16}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{15}{16} - \frac{\sqrt{33}}{16} = -1.297 s = -\frac{15}{16} + \frac{\sqrt{33}}{16} = -0.578
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.